Hooke’s Law Reviewed
F = !kx
Chapter 13
Vibrations and Waves
•
When x is positive
F is negative
;
,
•
When at equilibrium (x=0),
F = 0 ;
•
When x is negative
F is positive
;
,
1
2
Sinusoidal Oscillation
Graphing x vs. t
Pen traces a sine wave
A
T
A : amplitude (length, m)
T : period (time, s)
3
Phases
Some Vocabulary
x = A cos(! t " # )
= A cos(2$ ft " # )
% 2$ t
(
= A cos '
" #*
& T
)
4
Phase is related to starting time
f =
1
T
! = 2" f =
$ 2! t
'
x = A cos &
" #)
% T
(
2"
T
2! t 0
$ 2! (t " t 0 ) '
= A cos &
)( if # =
%
T
T
90-degrees changes cosine to sine
f = Frequency
! = Angular Frequency
T = Period
A = Amplitude
" = phase
#'
$
cos & ! t " ) = sin (! t )
%
2(
5
6
x
T
Velocity and
Acceleration vs. time
•
•
Velocity is 90°
“out of phase” with
x: When x is at max,
v is at min ....
Acceleration is 180°
“out of phase” with x
a = F/m = - (k/m) x
Find vmax with E conservation
1 2 1 2
kA = mvmax
2
2
k
vmax = A
m
v and a vs. t
x = A cos ! t
v = "vmax sin ! t
a = "amax cos ! t
v
T
Find amax using F=ma
!kx = ma
!kA cos " t = !mamax cos " t
k
amax = A
m
a
T
7
What is !?
8
Formula Summary
Requires calculus. Since
d
A cos ! t = "! Asin ! t
dt
k
vmax = ! A = A
m
!=
f =
1
T
! = 2" f =
x = A cos(! t " # )
v = "! Asin(! t " # )
k
m
a = "! 2 A(cos ! t " # )
2"
T
!=
k
m
9
Example13.1
10
Example 13.2
An block-spring system oscillates with an amplitude
of 3.5 cm. If the spring constant is 250 N/m and
the block has a mass of 0.50 kg, determine
A 36-kg block is attached to a spring of constant
k=600 N/m. The block is pulled 3.5 cm away from its
equilibrium positions and released from rest at t=0.
At t=0.75 seconds,
(a) the mechanical energy of the system
(b) the maximum speed of the block
a) what is the position of the block? a) -3.489 cm
(c) the maximum acceleration.
b) what is the velocity of the block? b) -1.138 cm/s
a) 0.153 J
b) 0.783 m/s
c) 17.5 m/s2
11
12
Example 13.4a
Example 13.3
An object undergoing simple harmonic motion follows the
expression,
x(t) = 4 + 2 cos[! (t " 3)]
A 36-kg block is attached to a spring of constant k=600
N/m. The block is pulled 3.5 cm away from its
equilibrium position and is pushed so that is has an initial
velocity of 5.0 cm/s at t=0.
Where x will be in cm if t is in seconds
The amplitude of the motion is:
a) 1 cm
b) 2 cm
c) 3 cm
d) 4 cm
e) -4 cm
a) What is the position of the block at t=0.75 seconds?
a) -3.39 cm
13
14
Example 13.4b
Example 13.4c
An object undergoing simple harmonic motion follows the
expression,
An object undergoing simple harmonic motion follows the
expression,
x(t) = 4 + 2 cos[! (t " 3)]
x(t) = 4 + 2 cos[! (t " 3)]
Here, x will be in cm if t is in seconds
Here, x will be in cm if t is in seconds
The period of the motion is:
a) 1/3 s
b) 1/2 s
c) 1 s
d) 2 s
e) 2/# s
The frequency of the motion is:
a) 1/3 Hz
b) 1/2 Hz
c) 1 Hz
d) 2 Hz
e) # Hz
15
16
Example 13.4d
Example 13.4e
An object undergoing simple harmonic motion follows the
expression,
An object undergoing simple harmonic motion follows the
expression,
x(t) = 4 + 2 cos[! (t " 3)]
x(t) = 4 + 2 cos[! (t " 3)]
Here, x will be in cm if t is in seconds
Here, x will be in cm if t is in seconds
The angular frequency of the motion is:
a) 1/3 rad/s
b) 1/2 rad/s
c) 1 rad/s
d) 2 rad/s
e) # rad/s
The object will pass through the equilibrium position
at the times, t = _____ seconds
a)
b)
c)
d)
e)
17
…,
…,
…,
…,
…,
-2, -1, 0, 1, 2 …
-1.5, -0.5, 0.5, 1.5, 2.5, …
-1.5, -1, -0.5, 0, 0.5, 1.0, 1.5, …
-4, -2, 0, 2, 4, …
-2.5, -0.5, 1.5, 3.5,
18
Simple Pendulum
Simple Pendulum
F = !mgsin "
x
x
sin " =
#
2
2
L
x +L
mg
F#!
x
L
F = !mgsin "
x
x
sin " =
#
2
2
L
x +L
mg
F#!
x
L
!=
g
L
" = " max cos(! t # $ )
Looks like Hooke’s law (k $ mg/L)
19
Simple pendulum
!=
20
Pendulum Demo
g
L
Frequency independent of mass and amplitude!
(for small amplitudes)
21
Damped Oscillations
Example 13.5
A man enters a tall tower, needing to know its height h.
He notes that a long pendulum extends from the roof
almost to the ground and that its period is 15.5 s.
(a) How tall is the tower?
22
In real systems,
friction slows motion
a) 59.7 m
(b) If this pendulum is taken to the Moon, where the
free-fall acceleration is 1.67 m/s2, what is the period
of the pendulum there?
b) 37.6 s
23
24
TRAVELING WAVES
Longitudinal (Compression) Waves
•Sound
•Surface of a liquid
•Vibration of strings
•Electromagnetic
•Radio waves
•Microwaves
•Infrared
•Visible
•Ultraviolet
•X-rays
•Gamma-rays
•Gravity
Sound waves are longitudinal waves
25
Compression and Transverse Waves Demo
26
Transverse Waves
Elements move perpendicular to wave motion
Elements move parallel to wave motion
27
Snapshot of a Transverse Wave
28
Snapshot of Longitudinal Wave
x
%
(
y = A cos ' 2! # $ *
& "
)
wavelength
%
x
%
(
y = A cos ' 2! # $ *
& "
)
x
y could refer to pressure or density
29
30
Moving Wave
x " vt
%
(
y = A cos ' 2!
" $*
&
)
#
Replace x with x-vt
if wave moves to the right.
Replace with x+vt if
wave should move to left.
Moving Wave: Formula Summary
+ #x
.
&
y = A cos - 2! % ! ft ( ) * 0
$
'
"
,
/
moves to right with velocity v
v = f!
Fixing x=0,
v
%
(
y = A cos ' !2" t ! $ *
&
)
#
f =
v
, v = f!
!
31
Example 13.6a
32
Example 13.6b
A wave traveling in the
positive x direction has a
frequency of f = 25.0 Hz as
shown in the figure. The
wavelength is:
A wave traveling in the
positive x direction has a
frequency of f = 25.0 Hz as
shown in the figure. The
amplitude is:
a)
b)
c)
d)
e)
a)
b)
c)
d)
e)
5 cm
9 cm
10 cm
18 cm
20 cm
5 cm
9 cm
10 cm
18 cm
20 cm
33
Example 13.7a
Example 13.6c
Consider the following expression for a pressure wave,
A wave traveling in the
positive x direction has a
frequency of f = 25.0 Hz as
shown in the figure. The speed
of the wave is:
a)
b)
c)
d)
e)
34
P = 60 ! cos ( 2x " 3t )
where it is assumed that x is in cm,t is in seconds and
P will be given in N/m2.
What is the amplitude?
a) 1.5 N/m2
b) 3 N/m2
c) 30 N/m2
d) 60 N/m2
e) 120 N/m2
25 cm/s
50 cm/s
100 cm/s
250 cm/s
500 cm/s
35
36
Example 13.7b
Example 13.7c
Consider the following expression for a pressure wave,
Consider the following expression for a pressure wave,
where it is assumed that x is in cm,t is in seconds and
P will be given in N/m2.
where it is assumed that x is in cm,t is in seconds and
P will be given in N/m2.
P = 60 ! cos ( 2x " 3t )
P = 60 ! cos ( 2x " 3t )
What is the wavelength?
a) 0.5 cm
b) 1 cm
c) 1.5 cm
d) # cm
e) 2# cm
What is the frequency?
a) 1.5 Hz
b) 3 Hz
c) 3/# Hz
d) 3/(2#) Hz
e) 3#/2 Hz
37
Example 13.7d
38
Example 13.8
Which of these waves move in the positive x direction?
Consider the following expression for a pressure wave,
1) y = !21.3" cos(3.4x + 2.5t)
2) y = !21.3" cos(3.4x ! 2.5t)
3) y = !21.3" cos(!3.4x + 2.5t)
4) y = !21.3" cos(!3.4x ! 2.5t)
5) y = 21.3" cos(3.4x + 2.5t)
6) y = 21.3" cos(3.4x ! 2.5t)
P = 60 ! cos ( 2x " 3t )
where it is assumed that x is in cm,t is in seconds and
P will be given in N/m2.
What is the speed of the wave?
a) 1.5 cm/s
b) 6 cm/s
c) 2/3 cm/s
d) 3#/2 cm/s
e) 2/# cm/s
7) y = 21.3" cos(!3.4x + 2.5t)
8) y = 21.3" cos(!3.4x ! 2.5t)
a)
b)
c)
d)
e)
5 and
1 and
5,6,7
1,4,5
2,3,6
6
4
and 8
and 8
and 7
39
Speed of a Wave in a Vibrating String
v=
40
Example 13.9
A string is tied tightly between points A and B
as a communication device. If one wants to
double the wave speed, one could:
T
m
where µ =
µ
L
a)
b)
c)
d)
e)
For different kinds of waves: (e.g. sound)
• Always a square root
• Numerator related to restoring force
• Denominator is some sort of mass density
41
Double the tension
Quadruple the tension
Use a string with half the mass
Use a string with double the mass
Use a string with quadruple the mass
42
Reflection – Fixed End
Superposition Principle
Reflected wave is inverted
Traveling waves can pass
through each other without
being altered.
y(x,t) = y1 (x,t) + y2 (x,t)
43
Reflection – Free End
Reflected pulse not inverted
45
44